Physics theoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Partition, Gibbs-entropy and free-energy factorization for independent finite systems

Physical scope

Finite composite models whose state set is explicitly the Cartesian product and whose microscopic energy is additive; no claim is made for interacting systems, interface terms, or experimental independence.

Statement

Let two finite microstate models from Finite statistical microstate model have nonempty state sets $Ω_{A}, Ω_{B}$ , energies $E_{A}, E_{B}$ in joules, and the same $T > 0$ . Assume their declared composite model has state set $Ω_{A} \times Ω_{B}$ and energy $E_{A B} (x, y) = E_{A} (x) + E_{B} (y)$ , with no interaction term. Under the canonical sampling assumption Canonical sampling assumption for finite and normalized statistical models, the partition function factorizes as $Z_{A B} = Z_{A} Z_{B}$ , the joint canonical law is $P_{A B} = P_{A} P_{B}$ , the Gibbs entropy satisfies $S_{A B} = S_{A} + S_{B}$ , and the statistical free-energy value satisfies $F_{A B} = F_{A} + F_{B}$ .

Facts & Assumptions

Given: Two finite nonempty state sets, additive composite energy, fixed $T > 0$ , and the canonical preparations of both subsystems and their composite.

[F1]

Finite statistical microstate model: each state set is finite and nonempty and the composite state set and additive energy are declared model data.

[F2]

Canonical sampling assumption for finite and normalized statistical models: for each model, $P (x) = e^{- β E (x)} / Z$ with $β = (k_{B} T)^{- 1}$ .

[F3]

Finite canonical partition, energy, variance, entropy and free-energy identities: $S = - k_{B} \sum_{x} P (x) \log P (x)$ and $F = \overset{ˉ}{E} - T S = - k_{B} T \log Z$ for each finite model.

Proof

Given: The finite product state set, additive energy, and the three canonical laws.

1.1F1F2algebra

For every $(x, y) \in Ω_{A} \times Ω_{B}$ , $e^{- β E_{A B} (x, y)} = e^{- β E_{A} (x)} e^{- β E_{B} (y)}$ . Reindexing the finite sum over the Cartesian product therefore gives $Z_{A B} = (\sum_{x} e^{- β E_{A} (x)}) (\sum_{y} e^{- β E_{B} (y)}) = Z_{A} Z_{B}$ . Dividing the product weight by $Z_{A} Z_{B}$ gives $P_{A B} (x, y) = P_{A} (x) P_{B} (y)$ .

2.1F1F2F3step 1.1algebra∎

Since each probability is positive, $\log P_{A B} (x, y) = \log P_{A} (x) + \log P_{B} (y)$ . Summing against $P_{A} (x) P_{B} (y)$ and using the finite normalizations yields $S_{A B} = S_{A} + S_{B}$ . The mean energy is ${\overset{ˉ}{E}}_{A B} = {\overset{ˉ}{E}}_{A} + {\overset{ˉ}{E}}_{B}$ by the same finite factorization, so $F_{A B} = {\overset{ˉ}{E}}_{A B} - T S_{A B} = F_{A} + F_{B}$ .

Partition, Gibbs-entropy and free-energy factorization for independent finite systems

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources